Problem #5 The figure at right shows the top view of a bar that can slide on two

Question

Problem #5 The figure at right shows the top view of a bar that can slide on two parallel rails without friction. A constant external force F app is applied to the bar causing it to slide to the right. A uniform magnetic field B is directed into the page. (a.) If the applied force causes the bar to slide with a constant velocity v = 2 m/s, what is the direction and magnitude of the current through the resistor? Take B = 0.5 T l = 10 cm, and R = 20 omega. Solve the problem algebraically first, then plug in numbers. (b.) Find the magnitude and direction of the magnetic force on the rod due to this motion. (If you do not solve part (a), give an algebraic answer in terms of well-identified quantities for partial credit.)

Explanation / Answer

B=0.5T

l=10cm=0.1m

R=20 ohm

cosider that the rod is forming a rectangular loop with the rail

Let's assume this is true, then the magnetic field induces an emf throughout the loop.
emf = -d(magnetic flux)/dt

For themagnetic flux, we have a uniform magnetic field and an area A, which will be area of rectangle

A = x*L
So flux is = BL *x

Now we take the first time derivative for the emf:
d/dt ( BLx )= BLdx/dt
Now, dx/dt and da/dt have the same rate of change, so dx/dt v so,
emf = - BLv
Let us assume direction is into the plane. So we get an emf of magnitude 1 BLv ongoing counter clockwise.

Now we have an emf going through the resistor as well, so from that we can calculate the current:
emf = IR
BLv/R = I This is part one.
I =0.5*0.1*2/20= 0.005A

b)

Magnetic force ,F=I(L*B)

so F=0.005*.1*0.5=0.00025N=2.5*10-4N

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